A thief starts his car from a point with an acceleration of 2m/s2. The police chasing the thief arrives at the point 10 s later and continue to chase the thief with a uniform velocity of 40 m/s in their car. In what time the police’s car will overtake the thief’s car.
Answers
Answer:
The Police took 7.5 seconds to reach 125 metres and catch the thief.
Explanation:
Thief's Car,
Acceleration (a) = 2 m/s²
Time (t) = 10s
Initial Velocity (u) = 0 m/s
According to the formula,
v = u+at
= 0+2*10
= 20 m/s
v²=u²+2aS .... Here, 'S' refers to distance travelled
= (20)² = (0)² + 2*2S
= 400 = 4S
= S = 100 m
Police's Car,
Initial Velocity (u) = 0 m/s
Final Velocity (v) = 40 m/s
Let's take the Distance as 100m.
v²=u²+2aS
= (40)²=(0)²+2*100a
= 1600 = 200a
= 8 m/s² = Acceleration
v = u+at
= 40 = 0+8t
= 40 = 8t
= t = 5s
The Police will travel the distance of 100m in 5 seconds.
But here, is the suspense. As the Police will reach 100 m, the thief would have covered more distance in 5 s.
Distance covered more by the thief = S = ut+0.5at²
= 0*5 + 0.5*2*5*5
= 25 m
Now, we have to find the time at which the police will also cover this more 25 metres.
S = ut+0.5at²
= 25 = 0.5*8*t²
= 25 = 4t²
= 6.25 = t²
= 2.5 s = t
So, by adding the time = (5+2.5) s
= 7.5 s
Hope this helps.